Answer:
Explanation:
turn over number = R max / [E]t = K2
From given , R max = 249 * 10 ^ -6 mol. L^-1
T [E]t = 2.23 n mol. L^-1
= 2.23 * 10^-9 mol. L^-1
Putting values in above equation,
= 111.65 * 10^3 S^-1
Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
Answer:
The answer to your question is all the formulas in bold has the same empirical formula
Explanation:
Data
Empirical formula CH₂O
Process
To solve this problem factor the subscripts of each formula and compare the result with the empirical formula given.
a) C₂H₄O₂ factor 2 2(CH₂O)
b) C₃H₆O₃ factor 3 3(CH₂O)
c) CH₂O₂ this formula can not be simplified
d) C₅H₁₀O₅ factor 5 5(CH₂O)
e) C₆H₁₂O₆ factor 6 6(CH₂O)
Total change in energy = Heat produced by system + work done on the system
= 228 - 55
= 173 kJ
Because this energy is released as heat, the reaction is exothermic.
